Why localisation has a cost
A hydrogen atom's electron is pulled towards the proton by the Coulomb force, and the closer it gets, the lower its potential energy. The classical atom collapsed for exactly this reason. So why doesn't the electron fall into the nucleus?
Bohr answered by decree: the lowest orbit is the lowest one allowed. Quantum mechanics gives a reason, and it is one of the most useful facts in physics. Confining a particle to a small region forces it to have a wide spread of momentum, and a spread of momentum costs kinetic energy. This essay makes that statement precise, as the uncertainty principle, then uses it to estimate the size of an atom in a few lines.
Building a particle out of waves
A wave with a single wavelength, $e^{ikx}$, has a definite momentum $\hbar k$. It also has the same density $|e^{ikx}|^2 = 1$ everywhere: a particle in that state could be anywhere at all. To describe a particle that is somewhere in particular, we need a wavefunction that is large in one region and small elsewhere, a wave packet. The only way to make one is to add together waves of many different wavelengths, arranged so that they reinforce in one region and cancel everywhere else.
This is a general fact about waves, not something special to quantum mechanics: a short pulse of sound contains a wide range of pitches, and a pure tone must last a long time. The narrower the packet, the wider the range of wavelengths it needs. And in quantum mechanics, each wavelength is a momentum. A particle confined to a narrow region is necessarily in a state made of many momenta, and a measurement of its momentum will give a correspondingly wide scatter of results.
What exactly is uncertain
To make this precise we need to say what "spread" means. Prepare a large number of particles in the same state. On some of them, measure the position; the results scatter around an average, and their standard deviation is $\Delta x$. On the others, measure the momentum, and call that standard deviation $\Delta p$. Both are properties of the state, fixed by its wavefunction:
$$\begin{aligned} \Delta x &= \sqrt{\langle x^2\rangle - \langle x\rangle^2},\\[4pt] \Delta p &= \sqrt{\langle p^2\rangle - \langle p\rangle^2}. \end{aligned}$$The uncertainty principle, in the form proved by Earle Kennard in 1927, says that for every possible state,
$$\Delta x\,\Delta p \;\ge\; \frac{\hbar}{2}.$$Three things about this statement are easy to get wrong. It is an inequality about standard deviations, with $\hbar/2$ on the right; rougher versions written with $h$ or "$\approx$" are estimates of the same idea, not different laws. It concerns the spreads of results over many identically prepared systems, not the error of any single measurement: an individual position measurement can be as sharp as the instrument allows. And it does not say the particle has a precise position and momentum that we are prevented from learning. No state exists in which both are sharp.
Diffraction is the same trade
The single slit shows the trade happening. Send electrons of momentum $p$ straight at a slit of width $w$. Passing through confines each electron's sideways position to within about $w$. Beyond the slit the electrons fan out in a diffraction pattern whose central peak spreads to an angle of about $\lambda/w$, which means each electron now has a sideways momentum anywhere in a range of about $p\lambda/w$. With $\lambda = h/p$ that range is $h/w$, and its product with the position range is about $h$. Narrow the slit to locate the electrons better, and the pattern widens in exact proportion.
Heisenberg's own first explanation, in 1927, was a thought experiment in which the light used to see an electron kicks it. Pictures of that kind, in which measurement disturbs, capture something real: measuring position does leave the momentum scattered. But the principle does not depend on any clumsiness in the apparatus. It holds for the state as prepared, before any measurement, because it follows from building a localised wavefunction out of waves. A better instrument cannot remove it.
Worked example: the size of an atom
Keep only the essentials, and aim for the right size rather than exact factors. Suppose the electron in hydrogen is confined within a distance about $r$ of the proton. Then its momentum must be spread over at least about $\hbar/r$, and since the average momentum is zero in a bound state, $\langle p^2\rangle$ is at least about $(\hbar/r)^2$. The energy is then roughly
$$E(r) \approx \frac{\hbar^2}{2m r^2} - \frac{e^2}{4\pi\varepsilon_0\, r}.$$The first term is the kinetic energy that confinement costs, growing as $1/r^2$. The second is the Coulomb energy, falling only as $1/r$. With $\hbar^2/2m = 0.0381$ eV nm² and $e^2/4\pi\varepsilon_0 = 1.440$ eV nm:
| $r$ (nm) | kinetic | Coulomb | total |
|---|---|---|---|
| 0.020 | +95.3 eV | −72.0 eV | +23.3 eV |
| 0.053 | +13.6 eV | −27.2 eV | −13.6 eV |
| 0.100 | +3.8 eV | −14.4 eV | −10.6 eV |
| 0.200 | +1.0 eV | −7.2 eV | −6.3 eV |
Squeezing in from 0.1 nm lowers the energy at first, but below about 0.05 nm the kinetic cost grows faster than the Coulomb gain, and at 0.02 nm the electron would not be bound at all. Setting $dE/dr = 0$ puts the minimum at
$$\begin{aligned} r &= \frac{\hbar^2}{m\,(e^2/4\pi\varepsilon_0)} = 0.053\ \text{nm},\\[4pt] E &= -13.6\ \text{eV}. \end{aligned}$$How much to trust it. These are the Bohr radius and hydrogen's true ground-state energy, but the exact agreement is partly luck: taking the momentum spread as $\hbar/2r$ instead of $\hbar/r$, equally defensible in an estimate, gives a radius four times smaller and an energy four times deeper. So this is an estimate of scale and mechanism, not a derivation of the ground-state energy. What it establishes firmly is the scale, about a tenth of a nanometre and about ten electronvolts, and the reason there is a bottom at all. The exact value comes from solving Schrödinger's equation for hydrogen, in a later essay.
Check: a dust grain. A dust grain of one microgram located to within one micrometre needs a velocity spread of only $\hbar/(2m\,\Delta x) = 5\times10^{-20}$ m/s, far below anything measurable. The principle applies to everything; for objects of everyday mass it never shows.
Optional: the general relation and its proof
The same argument works for any two quantities whose operators do not commute, as the previous essay showed position and momentum do not. For any state and any two Hermitian operators $\hat A$ and $\hat B$,
$$\Delta A\,\Delta B \;\ge\; \tfrac12\,\bigl|\langle[\hat A, \hat B]\rangle\bigr| .$$With $[\hat x, \hat p] = i\hbar$ the right side is $\hbar/2$, which is Kennard's relation.
Proof sketch. Shift each operator by its average, $\hat a = \hat A - \langle A\rangle$ and $\hat b = \hat B - \langle B\rangle$, so that $\Delta A^2 = \langle\psi|\hat a^2|\psi\rangle$ is the squared length of the state $\hat a|\psi\rangle$, and likewise for $\hat b$. The Cauchy–Schwarz inequality, that the overlap of two arrows is at most the product of their lengths, gives $\Delta A^2\,\Delta B^2 \ge |\langle\psi|\hat a\hat b|\psi\rangle|^2$. The number $\langle\hat a\hat b\rangle$ has imaginary part $\tfrac{1}{2i}\langle\hat a\hat b - \hat b\hat a\rangle = \tfrac{1}{2i}\langle[\hat A,\hat B]\rangle$, and a complex number's size is at least the size of its imaginary part. Taking square roots gives the relation.
Applied to the photon of the representations essay, with $\hat A$ and $\hat B$ the H/V and D/A questions, it says the same thing in two dimensions: a state sharp for one is maximally spread for the other.
Two questions
1. Why nuclear energies come in MeV. A proton inside a nucleus is confined to a region with $\Delta x \approx 2$ fm ($2\times10^{-15}$ m). Using $\Delta p \ge \hbar/2\Delta x$, estimate the smallest momentum spread and the kinetic energy that goes with it. The combination $\hbar c = 197$ MeV fm and the proton's rest energy $m_pc^2 = 938$ MeV make the arithmetic short.
Worked answer
$\Delta p\,c \ge \hbar c/2\Delta x = 197/4 \approx 49$ MeV. The kinetic energy is roughly $(\Delta p\,c)^2/2m_pc^2 = 49^2/1876 \approx 1.3$ MeV. Nuclear energies are indeed measured in MeV, against eV for atoms.
Kinetic energy from confinement goes as $\hbar^2/(m\,\Delta x^2)$, so size and mass both matter. The region is about 25,000 times smaller than an atom, which raises the energy by $25{,}000^2 \approx 6\times10^{8}$; the proton is about 1,800 times heavier than an electron, which lowers it by that factor. The net is a few hundred thousand, taking atomic electronvolts to nuclear MeV. The slip to avoid is scaling by size alone, which overshoots by the mass ratio, or forgetting that the size enters squared.
2. A better microscope. An engineer proposes an instrument precise enough to measure an electron's position to $10^{-15}$ m without disturbing its momentum at all. Would it beat the uncertainty principle? What exactly does the principle forbid?
Worked answer
It would not, even if it worked as described. The principle is not a limit on how precisely one measurement can be made. A single position measurement can be as sharp as you like. What the principle forbids is a state in which position and momentum are both sharp: if many electrons are prepared so that position measurements on them scatter by only $\Delta x$, then momentum measurements on others prepared the same way will scatter by at least $\hbar/2\Delta x$.
The tempting wrong turn is the disturbance picture: that the electron "really" has a sharp position and momentum and our instruments knock it about. The principle follows from what a localised wavefunction is made of, before any instrument is involved. An electron found within $10^{-15}$ m is, straight afterwards, in a state whose momentum spread is at least $\hbar/(2\times10^{-15}\ \text{m})$, however gently it was found.
Further reading
- Feynman, Lectures on Physics, Vol. III, §1-8 "The uncertainty principle" and Ch. 2, §2-2 "Measurement of position and momentum" and §2-4 "The size of an atom", which runs an estimate like the worked example.
- Born, Atomic Physics (Dover, 8th ed.), Ch. IV §7 "The Contradiction between the Wave Theory and the Corpuscular Theory, and its Removal": the single slit, Heisenberg's microscope and wave packets, with a rigorous proof in Appendix XII.
- Landau & Lifshitz, Quantum Mechanics (2nd ed.), §16 "Uncertainty relations"; Dirac, Principles of Quantum Mechanics (4th ed.), Ch. IV §24 "Heisenberg's Principle of Uncertainty".
Where this leaves us
The electron does not fall into the nucleus because it cannot be confined for free. A wavefunction squeezed into a region of size $\Delta x$ is built from momenta spread by at least $\hbar/2\Delta x$, and that spread costs kinetic energy growing as $1/\Delta x^2$, faster than the Coulomb attraction pays out. The balance falls at about a tenth of a nanometre and about ten electronvolts: the size and binding energy of atoms.
That estimate treated confinement loosely, as "within about $r$". Part II has now given us the tools to treat it exactly: write down the potential that holds a particle, solve for the stationary states, and read off the allowed energies as eigenvalues. The next essay does this for the simplest confinement there is, a particle in a box, and then asks what happens when the walls are not quite high enough to hold it.